Simplifying equations using boolean algebra

  1. Nov 19, 2007 #1
    1. Simplify the following equations using boolean algebra

    2. a) abc + ab'c'+ ab'c

    b) (abc)'+(a+c)'+b'c'


    3. Please help me to solve the above equations
     
  2. jcsd
  3. Nov 19, 2007 #2

    mjsd

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    pls first clarify your notations.
    I suspect abc means a AND b AND c?
    a+b means a OR b?
    a' means NOT a?
    right?

    now: what standard identities do u know?
     
  4. Nov 19, 2007 #3
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    ya. I could not use the complement notation while posting the query. I know all the standard notations. In second problem first i used sop and pos rules but got stuck up on next step. So please if u can help me in that.
     
  5. Nov 21, 2007 #4

    mjsd

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    sop? pos?
    what's that? anyway,

    have u tried playing around using the standard axioms:
    eg. associativity, commutativity, distributivity, De Morgan's Law, idempotence...etc.

    for example:
    a.a = a, a+a=a, (a.b).c = a.(b.c), a+b =b+a, a.(b+c)=a.b+a.c, a+(b.c)=(a+b).(a+c)
    a+a'=1, a.a'=0
    etc.
     
  6. Nov 21, 2007 #5
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    sop-sum of product rule
    and pos means product of sum rule.

    Ya i tried.
     
  7. Nov 22, 2007 #6

    mjsd

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    try reversing the distributive law to gather "common factor"
    repeat use of axioms etc. it is a bit of a trial and error process, unless you can see something in advance (which comes with experience only). But you can always check, at each step, that you have not make an error by checking the truth table for both the original and derived expression.
    another hint, sometimes it may even be useful to "add terms" into your expression


    of course, sometimes it may be difficult to tell whether you have reduced your expression as simple as possible.
     
  8. Nov 22, 2007 #7

    mjsd

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    another hint:
    x.y + y = y
    since
    x.y+y = (x+1).y = (1).y = y
     
  9. Nov 22, 2007 #8
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    I tried solving the first pb:
    solution:

    abc+ab'c'+ab'c
    =abc+a(b'+c')+ab'c (Product of Sum rule)
    =abc+ab'+ac'+ab'c
    =abc+ac'+ab'(1+c)
    =abc+ac'+ab' (1+c=1)
    =abc+a(c'+b')
    =abc+ab'c' (Sum of Product rule)

    Just check if the solution is correct
     
  10. Nov 23, 2007 #9

    mjsd

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    ?? b'c' => b'+c' ?

    is it b'c' or (b.c)' ?

    you can always check answer by simply writing out the truth table for the original expression and then compare with the one for the new expression.
     
  11. Nov 23, 2007 #10
    Oh ya u r correct.
    that's y i told u to check.
     
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